What is inverse oflaplace (e^{-s})/(s-2) ?

The answer to inverse oflaplace (e^{-s})/(s-2) is \H(t-1)e^{2(t-1)}

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inverse oflaplace (e^{-s})/(s-2)

inverse laplace

\frac{e ^{- s}}{s - 2}

H (t - 1) e^{2 (t - 1)}

Solution steps

L^{- 1} {\frac{e ^{- s}}{s - 2}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - 1) L^{- 1} {\frac{1}{s - 2}} (t - 1)

L^{- 1} {\frac{1}{s - 2}} : e^{2 t}

= H (t - 1) e^{2 (t - 1)}

\frac{d}{d x} (\frac{1}{x ^{2} ln ( x )})

f (x) = 7 x + 11, at x = 2

\frac{d y}{d x} = \frac{cos ( 8 x )}{e ^{8 y}}

g (x) = (5 - x)^{2} (2 x - 1)^{5}

\int_{- 8}^{10} \frac{x}{2} + 10 d x

What is inverse oflaplace (e^{-s})/(s-2) ?
The answer to inverse oflaplace (e^{-s})/(s-2) is \H(t-1)e^{2(t-1)}